Question 1.2: A cantilever beam of solid, circular cross-section supports ...

The direct loading system is equivalent to an axial load of 50 kN together with a bending moment of 50 \times 10^{3} \times 1.5 = 75,000 Nmm in a vertical plane. Therefore, at any point on the lower edge of the vertical plane of symmetry, there are compressive stresses due to the axial load and bending moment that act on planes perpendicular to the axis of the beam and are given, respectively, by Eqs. (1.2) and (15.9); that is,

\sigma= \underset{\delta A \rightarrow 0}{\lim} \frac{\delta P_{n}}{\delta A}  (1.2)

\sigma_{z}=\frac{M y}{I}  (15.9)

The shear stress, \tau_{x y}, at the same point due to the torque is obtained from Eq. (iv) in Example 3.1; that is,

\tau = \frac{TR}{J}   (iv)

The stress system acting on a two-dimensional rectangular element at the point is shown in Fig. 1.11. Note that, since the element is positioned at the bottom of the beam, the shear stress due to the torque is in the direction shown and is negative (see Fig. 1.8).

Again, \sigma_{n} and \tau may be found from first principles or by direct substitution in Eqs. (1.8) and (1.9). Note that \theta=30^{\circ}, \sigma_{y}=0, \text { and } \tau_{x y}=-28.3 N / mm ^{2}, the negative sign arising from the fact that it is in the opposite direction to \tau_{x y} in Fig. 1.8.

Then,

\sigma_{n}=\sigma_{x} \cos ^{2} \theta+\sigma_{y} \sin ^{2} \theta+\tau_{x y} \sin 2 \theta  (1.8)

\tau=\frac{\left(\sigma_{x}-\sigma_{y}\right)}{2} \sin 2 \theta-\tau_{x y} \cos 2 \theta  (1.9)

\tau=(-21.2 / 2) \sin 60^{\circ}+28.3 \cos 60^{\circ}=5.0 N / mm ^{2} (acting in the direction AB)

Different answers are obtained if the plane AB is chosen on the opposite side of AC.